Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If $\angle$ A=$\pi /2$, then $\tan \left(C/2\right)$is equal to?

• A. $\frac{a-c}{2b}$
• B. $\frac{a-b}{2c}$
• C. $\frac{a-c}{b}$
• D. $\frac{a-b}{c}$

Answer 1

The correct option is D $\frac{a-b}{c}$

$\tan C=\frac{c}{b}$

$\tan C=\frac{2\tan \frac{c}{2}}{1-\tan {}^2\frac{c}{2}}$ $[\because \tan 2A=\frac{2\tan A}{1-\tan {}^{2}A}]$

$\therefore \frac{c}{b}=\frac{2\tan \frac{c}{2}}{1-\tan {}^2\frac{c}{2}}\Rightarrow c-c\tan {}^2\frac{c}{2}-2b\tan \frac{c}{2}=0$

Solving above quadratic equation:

$\tan \frac{c}{2}=\frac{-2b\pm \sqrt{4b^2+4c^2}}{2c}$

$\tan \frac{c}{2}=\frac{-2b\pm 2\times a}{2c}$

$\tan \frac{c}{2}=\frac{a-b}{c}or\frac{-a-b}{c}$

Hence, $\frac{a-b}{c}$ is the correct answer.